Physics midterm question: comparing vertical forces supporting tilted beams

Physics 205A Midterm 2, fall semester 2019
Cuesta College, San Luis Obispo, CA

A force F₁ pulls up at the end of a uniform beam to hold it stationary at an angle of 80° above the horizontal, and a force F₂ pulls up at the end of an identical uniform beam to hold it stationary at an angle of 10° above the horizontal. (Calculate all torques with respect to the pivot, located at the base of the beams.) Discuss why these forces 
 F₁ and F₂ have the same magnitude. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. the magnitude of the weight force w is the same for both higher and lower beams; and
  2. for each beam, the lever arm for the applied force F is twice the lever for the weight force w (2⋅ℓ_w = ℓ_F); and
  3. Newton's first law for rotations applies to both higher and lower beams, where the ccw force torque F⋅(ℓ_F) and cw weight torque w⋅(ℓ_w) must balance each other out, and so: F⋅(ℓ_F) = w⋅(ℓ_w), F = w⋅(ℓ_w/ℓ_F) = w⋅(ℓ_w/(2⋅ℓ_w)) = w/2; such that
  4. the applied forces on the higher and lower beam must be equal in magnitude, as they are both equal to one-half of the weight of the beam.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Does not explicitly note that the ℓ_F lever arm is always twice the ℓ_w lever arm for both situations. Instead, argues that since the ℓ_F and ℓ_w values for the higher beam are both bigger than the respective ℓ_F and ℓ_w values for the lower beam, then the higher beam F = w⋅(ℓ_w/ℓ_F) = w⋅(bigger/bigger) must be equal to the lower beam F = w⋅(ℓ_w/ℓ_F) = w⋅(smaller/smaller), but only implicitly demonstrates how the "bigger/bigger" ratio is exactly equal to the "smaller/smaller" ratio by use of a scaled drawing instead of using geometry/trigonometry, etc.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. As (r), but does not clearly/correctly show ℓ_F and ℓ_w lever arms for both situations. At least has two sets of Newton's first law for rotations, one for the higher beam and one for the the lower beam, setting the ccw torques equal to the cw torques.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Newton's first law to torques, forces, and perpendicular lever arms.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02sQm5
p: 3 students
r: 18 students
t: 14 students
v: 14 students
x: 3 students
y: 0 students
z: 0 students

A sample "p" response (from student 5281):

Physics quiz question: comparing torques on supported square

Physics 205A Quiz 5, fall semester 2019
Cuesta College, San Luis Obispo, CA

A uniform 0.40 m × 0.40 m square with 
a mass of 0.50 kg is pivoted at one corner. It is supported by a 45° diagonal force at the opposite corner such that the bottom edge is parallel to the ground. Calculate all torques with respect to the corner pivot. The torque exerted by the __________ has a greater magnitude. (A) weight force.
(B) support force.
(C) (There is a tie.)
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (C)

Since the square is in static equilibrium ("is supported") and does not rotate, then the net torque on it is equal to zero (Στ = 0), such that the counterclockwise torque of the weight force on the square must equal the clockwise torque of the support force on the square:

(ccw) τ_w = (cw) τ_Fsupport.

Sections 70854, 70855
Exam code: quiz05Gu1L
(A) : 23 students
(B) : 21 students
(C) : 7 students
(D) : 0 students

Success level: 13%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.24

Physics quiz question: weight torque on supported square

Physics 205A Quiz 5, fall semester 2019
Cuesta College, San Luis Obispo, CA

A uniform 0.40 m × 0.40 m square with 
a mass of 0.50 kg is pivoted at one corner. It is supported by a 45° diagonal force at the opposite corner such that the bottom edge is parallel to the ground. Calculate all torques with respect to the corner pivot. The magnitude of the torque exerted by the weight force is:
(A) 0.98 N·m.
(B) 1.4 N·m.
(C) 2.0 N·m.
(D) 2.8 N·m.

Correct answer (highlight to unhide): (A)

The weight force w of Earth acting on the beam acts at the center of gravity, directly downwards with a magnitude m·g. The perpendicular lever arm ℓ for the weight force on the beam must extend from the pivot point to perpendicularly intercept the weight force line of action (which lies along the weight force vector), such that this will be a horizontal line that is half the side of the square:

ℓ = 0.20 m.

The magnitude of the (counterclockwise) torque exerted by the weight force on the beam is then:

τ = w·ℓ,

τ = (m·g)·(0.20 m),

τ = ((0.50 kg)·(9.80 m/s²))·(0.20 m),

τ = 0.98 N·m.

(Response (B) is m·g·(0.20 m)/sin(45°); response (C) is m·g·(0.40 m); and response (D) is m·g·(0.40 m)/sin(45°).)

Sections 70854, 70855
Exam code: quiz05Gu1L
(A) : 26 students
(B) : 16 students
(C) : 6 students
(D) : 4 students

Success level: 50%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.65

Physics quiz archive: rotations, torque, pressure, buoyancy, fluid flow

Physics 205A Quiz 5, fall semester 2019
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, version 1
Exam code: quiz05Gu1L



Sections 70854, 70855 results

0- 6 :	** [low = 6]
7-12 :	*********
13-18 :	***********
19-24 :	***************** [mean = 19.8 +/- 6.3]
25-30 :	************* [high = 27]

Online reading assignment: torque and rotations

Physics 205A, fall semester 2019
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters and previewing a presentation on torque and rotations.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"Magnitude of torque is equal to the magnitude of force times the lever arm, which means there is more torque when you push near the outside end of lever in the direction of the hinge. The line of action is parallel and overlapping the force applied."

"How to solve for torque; it can vary due to the position of the force and the length of an object that it accompanies. if a force is applied at the very end of a length while being perpendicular to the object you'll achieve maximum torque."

"The idea of Newtons's first and second law when applied to torque and rotations. Like force, when net torque is equal to zero, Newton's first law is in effect; moreover, if torque adds up to something other than zero, Newton's second law applies."

"The lug nuts do get supertight so it was helpful to see an example that I could personally relate to which makes torque and twist easier to understand. The force and direction (ccw/cw) also made sense because of the rotation or 'twist' and the force making it do so."

"The line of action and the lever arm are important concepts when talking about torque. The line of action being an extended line drawn co-linear with the force and the lever arm referring to the perpendicular distance between the line of action and the axis of rotation."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"Exactly how to apply Newton’s laws to the rotational motion."

"The line of action and the perpendicular lever arm. Although I generally understood the examples of Newtons's laws in relation to torque, I need help in class--through examples--to see how these lines of actions are drawn and how they allow the determination of a lever arm or vice versa."

"I think I understand torque for the most part. I just have trouble visualizing how the line of action and lever arm work for different/more complicated real-world examples."

"Just the formulas obviously, but nothing practice won't help."

"I am confused about how torque can zero, or would that be considered a non-existent idea?"

"I am having a hard time understanding the equilibrium of a rigid body. I understand that if a rigid body is in equilibrium the sum of the externally applied forces is zero and the sum of the applied toques is zero, but what if it isn't in equilibrium, and how do we know if it is, and will we even have to worry about that?"

What is the SI (Système International) unit for torque?

"A newton meter (N·m)."

"Newton·meters, not to be confused with joules."

Briefly describe how the line of action should be drawn for a given force.

"The line of action is drawn collinear with force."

"Extend a line along force vector."

"The line of action should be drawn co-linear with the force. It is an extended line and is usually dotted."

When a lever arm (or moment arm) is drawn, briefly explain where it starts, and how it should intersect the line of action for a force.

"The lever arm is defined as the perpendicular distance from the axis rotation to the line of action of the force."

"Starts at the axis of rotation origin, and is perpendicular to the line of action."

"Extend a line along this force vector--this is the line of action, which is just a guide to find the perpendicular lever arm, which starts at the pivot point, and at the other end must intersect the line of action perpendicularly."

For the stuck wrench (assuming that it is rigid and not flexing), Newton's rotational __________ law applies, and the clockwise torques and counterclockwise torques acting on the wrench are __________.

first; balanced.	********************************************* [45]
second; unbalanced.	**** [4]
(Unsure/lost/guessing/help!)	[0]

For the crane, Newton's rotational __________ law applies, and the clockwise torques and counterclockwise torques acting on the crane are __________.

first; balanced.	** [2]
second; unbalanced.	*********************************************** [47]
(Unsure/lost/guessing/help!)	[0]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"Can all the forces besides weight force be considered in calculating torque?" (Yes, definitely.)

"How would the guy use the spanner to apply a greater amount of torque? Use a longer spanner?" (Yes, or apply a perpendicular force to end of the original size spanner.)

"Am I just completely missing the section of Newton's rotational laws? I am assuming we are just applying the laws we already know?"

"Are Newton's rotational laws the same as his laws of motion?" (The textbook does not actually state Newton's laws as applied to rotations, but Newton's first law for rotations is when net torque is zero if rotational motion is constant (which includes no rotations), and Newton's second law for rotations is when the net force is non-zero, such that rotational motion will be changing. Newton's third law for rotations is very obscure; so we're not going there.)

Your use of 'esoteric' had a double meaning in that few people would use such a means to calculate torque (line of action and perpendicular lever arm) would be equaled by the number of people who would understand the meaning of 'esoteric.' LOL"

"I've never seen a crane fall." (It is bad luck to see a crane fall.)

Physics midterm question: comparing horizontal forces supporting tilted beams

Physics 205A Midterm 2, fall semester 2018
Cuesta College, San Luis Obispo, CA

A horizontal force is applied to hold a uniform beam stationary at an angle of 80° above the horizontal, and another horizontal force is applied to hold it stationary at an angle of 10° above the horizontal. (Calculate all torques with respect to the pivot, located at the base of the beam.) Discuss why less force required to hold the beam when it is at the higher 80° angle. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. the magnitude of the weight force w is the same for both higher and lower beams; but
  2. the lever arms ℓ_F are not the same, where ℓ_F is longer for the higher beam, and shorter for the lower beam; and
  3. the lever arms ℓ_w are not the same, where ℓ_w is shorter for the higher beam, and longer for the lower beam; and
  4. Newton's first law for rotations applies to both higher and lower beams, where the ccw force τ = F⋅ℓ_F and cw weight τ = w⋅ℓ_w must balance each other out, and so:

     F⋅ℓ_F = w⋅ℓ_w,
     
     F = w⋅(ℓ_w/ℓ_F);

     such that
  5. for the higher beam, the shorter ℓ_w in the numerator and longer ℓ_F in the denominator means that the applied force F is smaller than for the lower beam (where it has a longer ℓ_w in the numerator and a shorter ℓ_F in the denominator).
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. At least demonstrates that for the higher beam ℓ_w is shorter and ℓ_F is longer, but typically discusses only how one of these contributes to making the applied force smaller.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Typically argues that F is smaller for the higher beam because ℓ_F is larger (while claiming ℓ_w is the same for both beams); or F is smaller for the higher beam because ℓ_w is smaller (while claiming ℓ_F is the same for both beams).
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Newton's first law to torques, forces, and perpendicular lever arms.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02r3iN
p: 17 students
r: 4 students
t: 21 students
v: 10 students
x: 5 students
y: 0 students
z: 0 students

A sample "p" response (from student 5250):

Physics quiz question: falling beam torque

Physics 205A Quiz 5, fall semester 2018
Cuesta College, San Luis Obispo, CA

A uniform beam with a 0.15 kg mass and 1.2 m length pivots at one end as it falls over starting from an angle of 5° from the vertical. Calculate all torques with respect to the bottom of the beam. While at its starting point of 5°, the magnitude of the torque exerted by the weight force on the beam is:
(A) 0.077 N·m.
(B) 0.88 N·m.
(C) 1.5 N·m.
(D) 1.8 N·m.

Correct answer (highlight to unhide): (A)

The weight force w of Earth acting on the beam acts at the center of gravity, directly downwards with a magnitude m·g. The perpendicular lever arm ℓ for the weight force on the beam must extend from the pivot point to perpendicularly intercept the weight force line of action (which lies along the weight force vector), such that this will be a horizontal line of length:

ℓ = (L/2)·cos(85°).

The magnitude of the (clockwise) torque exerted by the weight force on the beam is then:

τ = w·ℓ,

τ = (m·g)·(L/2)·cos(85°),

τ = ((0.15 kg)·(9.80 m/s²))·((1.2 m)/2)·cos(85°),

τ = 0.0768713651... N·m,

or to two significant figures, the torque of the weight force on the beam has a magnitude of 0.077 N·m.

(Response (B) is (m·g)·(L/2)·sin(85°); response (C) is m·g; and response (D) is m·g·L.)

Sections 70854, 70855
Exam code: quiz05Ro74
(A) : 6 students
(B) : 11 students
(C) : 4 students
(D) : 31 students

Success level: 12%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.25

Physics quiz archive: rotations, torque, pressure, buoyancy, fluid flow

Physics 205A Quiz 5, fall semester 2018
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, version 1
Exam code: quiz05Ro74



Sections 70854, 70855 results

0- 6 :
7-12 :	******** [low = 9]
13-18 :	**************
19-24 :	************************* [mean = 19.6 +/- 5.2]
25-30 :	***** [high = 30]

Online reading assignment: torque and rotations

Physics 205A, fall semester 2018
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters and previewing a presentation on torque and rotations.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"I seem to have a pretty good understanding of the new formulas for torque, as well I seem to understand how to apply them."

"That torque can be counterclockwise or clockwise. Also that when net torque is not zero then an object will rotate but when it is zero then the object's rotational motion will not change."

"For Newton's first law for rotation forces means the torques add up to zero, such that its rotational motion is constant. Newton's second law for rotation forces means the net torque is non-zero, and the rotational motion is not constant."

"How Newton's first and second laws apply to torque and rotation. Newton's first law says that when the counterclockwise torques are balanced by clockwise torques on that same object the object is either stationary or in constant rotational motion. Newton's second law says if the counterclockwise and clockwise torques on an object don't balance, then the net torque can be found by summing the counterclockwise (positive) torques and clockwise (negative) torques, and the object's rotational motion will be changing."

"A lever arm is perpendicular distance from the axis of rotation to the line of action along which the force acts. Also torque is the product of force and lever arm."

"The explanation of torque with regards to opening a door. What I got from the example was that torque depends on the magnitude of the force and on the distance from the point where the force is applied relative to to the axis of rotation. In this example of the door, the hinge is the axis of rotation. The closer your force applied is to the hinge, the harder it is to open the door. But the further away you are, for example the door knob, the easier it will be to open the door."

"Torque is the the twisting force causing rotation. Torque is the product of force and the perpendicular lever arm. The force vector line of action is perpendicular to the lever arm."

"Torque is to twist and to calculate it you need to multiply force times the perpendicular lever arm."

"The presentation was very straightforward."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"Exactly how the torque is measured/calculated is not totally intuitive to me. If I were to work through more examples, I think I could get a concrete understanding of how this force is calculated and applied."

"I don't understand rotational equilibrium."

"I'm not sure how to set up the torque diagrams."

"How a perpendicular lever arm plays a role. I understand that it is multiplied by the force to find torque, but what exactly is it?"

"I was a little confused about how we determine the line of action from just looking at a model or picture of a scenario."

"Torque = twisting force?"

"I'm still confused on what exactly torque is. I read through the presentation twice and discussed with other friends what I thought it may be. It's just a 'twisting' force from what I'm gathering. I'll have to spend more time digging into it."

"I don't understand the concept torque overall :("

"I'm not quite sure how to identify whether/how the net torque on an object should add up to zero."

"I didn't really feel confident in this reading and I found most of the material confusing."

"I did not get to this yet."

What is the SI (Système International) unit for torque?

"Torque is measured in newton meters."

"N·m (BUT NOT THE SAME AS JOULES)."

Briefly describe how the line of action should be drawn for a given force.

"The line of action should extend along the force vector."

"It is the line through the point at which the force is applied in the same direction as the force vector."

"I will get back to you on this."

"I did not get to this yet."

When a lever arm (or moment arm) is drawn, briefly explain where it starts, and how it should intersect the line of action for a force.

"It starts at the pivot point and must intersect perpendicularly with the line of action."

"The line that creates a 90° angle between the axis of rotation and the line of action."

"The lever arm is drawn perpendicular to the line of action and connects to the pivot point on the object."

"Not too sure."

For the stuck wrench (assuming that it is rigid and not flexing), Newton's rotational __________ law applies, and the clockwise torques and counterclockwise torques acting on the wrench are __________.

first; balanced.	************************************** [38]
second; unbalanced.	** [2]
(Unsure/lost/guessing/help!)	*** [3]

For the crane, Newton's rotational __________ law applies, and the clockwise torques and counterclockwise torques acting on the crane are __________.

first; balanced.	** [2]
second; unbalanced.	************************************** [38]
(Unsure/lost/guessing/help!)	*** [3]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"I don't understand these examples."

Can you do more questions in class?"

"I would like to go over lines of action and lever arms."

"If the mechanic had a longer wrench would it make it easier to rotate the nut?" (Yes, provided that he also exerted a force perpendicular to that longer wrench.)

"Will the line of action and perpendicular lever arm always form what appears to be a right triangle with the force?" (No--while the lever arm will always intersect the line of action at right angles, they may not necessarily have the force as a "hypotenuse." We'll go over some examples of this in class.)

"Torque is positive when the force produces a counterclockwise rotation around an axis, and torque is negative when the force produces a clockwise rotation. How is this determined and why?" (This rotational direction is the same as for angles measured around the unit circle: going counterclockwise from the +x axis is a positive angle, and going clockwise from the +x axis is a negative angle.)

"It not as easy as balanced and not balanced for torques, is it?" (That's essentially the difference between Newton's first law for rotations (ccw torques and cw torques cancel out) and Newton's second law for rotations (ccw torques do not cancel cw torques).)

"For the problems where object 1 collides into a stationary object 2 such that they stick together afterwards (completely inelastic collision), this equation seems after to solve since we can instead plug in the known numbers right away:

(m₁ + m₂)·v_f = m₁·v₀₁ + 0.

For momentum conservation, can we use this equation every time instead of always starting from ΣF·∆t = ∆p₁ + ∆p₂?"

(That would be okay, if the only type of collisions you ever use that equation for are always (1) completely inelastic (sticking together afterwards, such that v_f1 = v_f2 = v_f); and (2) have a stationary object 2 (such that v₀₂ = 0). However, (1) not every collision you will need to solve will be completely inelastic (although so far that's been the case); and (2) not every collision you will ned to solve will have a stationary object 2 (such as the Mission: Impossible 2 stuntmen, and the airliner colliding with the turkey buzzard). So in order to handle any type of collision with any given initial object velocity, I think you should go with the basic ΣF·∆t = ∆p₁ + ∆p₂ equation that will be given on the quiz and the midterm equation sections.)

"We should start Midterm 2 preparation two weeks before the actual midterm date. Or a week-and-a-half before? Or at least one week before? There's got to be some way you can alter the calendar to fit in earlier preparation time." (Well, like the first midterm, you'll be given a study guide and a list of sample midterm questions a week before. At this point the best ways you can start studying for the midterm now is to (1) study for each quiz as it comes; and (2) carefully go over each of the example midterm problems that are periodically shown in class or assigned for homework. Other than that, that midterm hasn't even been written yet, so I feel that keeping up with the current material right now is the best way to build a basic preparatory knowledge of what will eventually show up on Midterm 2.)

Physics midterm question: tension in cable lowering a boom crane

Physics 205A Midterm 2, fall semester 2017
Cuesta College, San Luis Obispo, CA

A cable anchored to a wall suspends a uniform beam, which can either be held 40° above the horizontal, or 40° below the horizontal. (Calculate all torques with respect to the pivot, located at the base of the beam.) Discuss why there is less tension in the cable when the beam is held 40° above the horizontal, compared to when it is held 40° below the horizontal. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. for either case (40° above or below the horizontal), the ccw cable tension torque = ℓ_T⋅T and cw weight torque = ℓ_w⋅w; and
  2. since Newton's first law for rotations applies to both cases, the ccw cable tension torque equals the cw weight torque such that ℓ_T⋅T = ℓ_w⋅w; and
  3. the perpendicular lever arm ℓ_w and the weight force is the same for both cases, where ℓ_w = (L/2)·cos40° and w = m⋅g; and
  4. since the cw weight torque is the same for both cases, the ccw cable tension torque is also the same for both cases; and
  5. the first case (40° above the horizontal) has a larger perpendicular lever arm ℓ_T, such that its tension T has a smaller magnitude compared to the second case (40° below the horizontal), with a smaller perpendicular lever arm ℓ_T, such that its tension T has a greater magnitude. Must at least clearly draw these ℓ_T lever arms to compare their relative lengths.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May have incorrect or missing cable lever arm.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Has at least three of the (1)-(5) steps above.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Newton's first law to torques, forces, and perpendicular lever arms.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02bu2Z
p: 14 students
r: 3 students
t: 20 students
v: 10 students
x: 3 students
y: 0 students
z: 0 students

A sample "p" response (from student 2635):

Physics quiz question: comparing lever arms for loaded beam forces

Physics 205A Quiz 5, fall semester 2017
Cuesta College, San Luis Obispo, CA

A uniform beam with a mass of 10 kg and length 2.0 m has a 3.0 kg load hanging from its end, and is suspended by a horizontal cable attached to a wall. (Calculate all torques with respect to the pivot.) The __________ force has the longest perpendicular lever arm.
(A) horizontal cable tension.
(B) beam weight.
(C) load.
(D) (There is a tie.)

Correct answer (highlight to unhide): (A)

The weight of the boom acts at its center of gravity, straight downwards. The perpendicular lever arm ℓ_w for the weight force w extends from the pivot to perpendicularly intercept the weight force line of action, such that this will be a horizontal line of length:

ℓ_w = (1.0 m)·cos(70°) = 0.34 m.

Similarly the perpendicular lever arm ℓ_load for the load force F_load extends from the pivot to the perpendicularly intercept the load force of action, such that:

ℓ_load = (2.0 m)·cos(70°) = 0.68 m.

The perpendicular lever arm for the horizontal cable force extends from the pivot point to perpendicularly intercept the tension force of action, such that:

ℓ_cable = (2.0 m)·sin(70°) = 1.9 m.

Sections 70854, 70855
Exam code: quiz05nWaW
(A) : 21 students
(B) : 11 students
(C) : 14 students
(D) : 3 students

Success level: 43%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.53

Physics quiz archive: rotations, torque, pressure, buoyancy, fluid flow

Physics 205A Quiz 5, fall semester 2017
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, version 1
Exam code: quiz05nWaW



Sections 70854, 70855 results

0- 6 :
7-12 :	***** [low = 9]
13-18 :	***************
19-24 :	********************* [mean = 19.7 +/- 4.8]
25-30 :	***** [high = 27]

Online reading assignment: torque and rotations

Physics 205A, fall semester 2017
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters and previewing a presentation on torque and rotations.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"Torque is a twisting motion that is caused by a force on a perpendicular lever arm. The best example from the presentation that made complete sense to me was the lugnut being removed."

"The SI unit for torque is newton times meter."

"From reading the material, I understand how perpendicular lever arms work and to sketch them in a free body diagram."

"Newton's first law of motion also applies to torque. If the counterclockwise and clockwise torques balance out to either be in equilibrium or at a constant motion it can be covered by Newton's first law."

"Torque is known as the twisting motion that a force exerts on an object.It can have either a clockwise motion of rotation or a counter clockwise motion of rotation. With Newton's first law, the clockwise and counterclockwise torques are balanced and are therefore able to be equal to each other."

"I feel I understand how to setup a torque diagram with the line of action and a perpendicular lever arm. I also feel I can identify if a certain situation applies to Newton's first or second law."

"Newton's first law of rotation states that the ccw torque and the cw torque balance, therefore the net torque equals zero. And for Newton's second law of rotation, net torque is equal to a non-zero."

"I understand the general introduction of torque. I see how to use the equation."

"Torque is honestly a really hard concept and would really appreciate some help."

"There wasn't a lot I understood. I got what the definition of a torque is."

"I do not understand torque."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"I am having trouble with understanding the line of action concept. I am not sure what it exactly means by line of action of a force."

"The perpendicular lever arms--I'm assuming we'll go over this in class and that should help."

"I have to re-examine that line of action thing. I really skimmed through the page though. I am sorry."

"How to determine the pivot point of an object. Also I think I would benefit from further explanation to some of the thought processes in determining between whether Newton's first law or second law applies to rotations."

"Rigid objects in equilibrium. If you could go over an example that would be nice."

"All of it."

"I'm super-confused. I get all of the terms I’m just not understanding how they all work with one another. Any example would be nice."

"I did not find the material confusing. I found everything to be pretty straightforward."

"I actually find this to be relatively intuitive, based on what we've already learned about Newton's laws. I also feel like it makes sense based on what I observe in life. Can't really think of anything I am confused by here."

"Most of it was pretty clear."

"Nothing, for now."

What is the SI (Système International) unit for torque?

"It is newtons times meters."

"N·m."

"Newton meter? I got this answer from a Google search. The blog specified that it is NOT joules, but I don't think the kind of SI unit was named."

"It is force times distance?"

"Joules?"

Briefly describe how the line of action should be drawn for a given force.

"Co-linear to the force."

"It should continue in the direction of the force."

"Extended from the direction of force."

"Drawn with a dashed line."

"Straight?"

"A counter-clockwise or clockwise direction?"

"Perpendicular to the ground or the force exerted on it?"

"I'm lost."

When a lever arm (or moment arm) is drawn, briefly explain where it starts, and how it should intersect the line of action for a force.

"It starts at the pivot point of the object in question, and it intersects the line of action at a perpendicular angle."

"The lever arm should start at the rotation axis and is perpendicular to the line of action."

"The lever arm should start at the pivot point. Its other end must intersect the line of action perpendicularly."

"I'm not sure--this is where I get confused."

"I'm lost."

"It starts in the center of the circle and goes outwards?"

"Starts at center of mass?"

For the stuck wrench (assuming that it is rigid and not flexing), Newton's rotational __________ law applies, and the clockwise torques and counterclockwise torques acting on the wrench are __________.

first; balanced.	************************************ [36]
second; unbalanced.	*** [3]
(Unsure/lost/guessing/help!)	***** [5]

For the crane, Newton's rotational __________ law applies, and the clockwise torques and counterclockwise torques acting on the crane are __________.

first; balanced.	* [1]
second; unbalanced.	************************************** [38]
(Unsure/lost/guessing/help!)	***** [5]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"When I should use sines and cosines? I know cosine is for the x-components and sines is for y-components, but I don't feel like it is as simple as that." (It is not. But just remember to use SOH-CAH-TOA, no matter how weird the triangle looks, and you should be okay.)

"I really liked how you explained each part of the momentum conservation equation in Monday’s class. It really helped. This topic however is overwhelming for me."

"So the longer the lever arm, the greater force needed to rotate the object? (No, the opposite of that. If you have more "leverage," then you can apply less force to cause the same torque.)

"I could seriously benefit from lecturing on this in class instead of trying to understand it by reading about it."

"I think it might just be how the textbook words things, but I can't understand torque very well. Please go over this in class."

"I was pretty lost on this, this is my first time seeing this stuff!"

"I need to put my nose in the book more."

Physics final exam question: comparing boom crane piston forces

Physics 205A Final Exam, fall semester 2016
Cuesta College, San Luis Obispo, CA

A telescoping boom crane can either be supported by two types of diagonal pistons, attached to the same point along the boom. (Calculate all torques with respect to the pivot, located at the base of the boom, approximated here as a uniform beam.) Determine which piston is exerting a greater magnitude force on the boom, or if there is a tie. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. for either case (piston (A) or piston (B)), the cw piston torque = l_piston⋅F_piston and ccw weight torque = l_weight⋅w; and
  2. since Newton's first law applies to both cases (piston (A) or piston (B)), the ccw torque of the piston on the boom equals the cw torque of the weight on the boom such that l_piston⋅F_piston = l_weight⋅w; then
  3. for piston (A) and piston (B), the ccw torque of the piston on the boom l_piston⋅F_piston has the same value;
  4. the perpendicular lever arm l_piston is smaller for piston (A) and larger for piston (B), and since l_piston⋅F_piston remains constant, then F_piston must be larger for piston (A) and smaller for piston (B).
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May have only implied Newton's first law in order to compare (equal) piston torques.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. May not have recognized difference in l_piston lever arms (claiming that they are the same, and thus F_piston forces are the same); or recognizes difference in l_piston, but claims that piston (B) has the greater F_piston force.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Newton's first law to torques, forces, and perpendicular lever arms. May have pistons exerting different amounts of torques.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finali0w4
p: 22 students
r: 3 students
t: 19 students
v: 5 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 6436):

Physics midterm question: extended boom crane torques, forces

Physics 205A Midterm 2, fall semester 2016
Cuesta College, San Luis Obispo, CA

A telescoping boom crane extends its length (while keeping the mass of the boom constant), supported by a vertical piston. (Calculate all torques with respect to the pivot, located at the base of the boom, approximated here as a uniform beam.) Discuss why the force exerted by the piston increases as the boom is extended. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. for either case (short boom or long boom), the cw piston torque = ℓ_piston⋅F_piston and ccw weight torque = ℓ_w⋅w; and
  2. since Newton's first law applies to both cases (short boom and long boom), the ccw torque of the piston on the boom equals the cw torque of the weight on the boom such that ℓ_piston⋅F_piston = ℓ_w⋅w; then
  3. for the long boom case compared to the short boom case, the perpendicular lever arm ℓ_w is longer (while w remains the same), and since ℓ_piston remains the same, then F_piston must increase.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Newton's first law to torques, forces, and perpendicular lever arms.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: midterm02oPt0
p: 17 students
r: 10 students
t: 11 students
v: 17 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student):

Physics quiz question: piston force on boom crane

Physics 205A Quiz 5, fall semester 2016
Cuesta College, San Luis Obispo, CA

A boom crane is supported by a vertical piston. (Calculate all torques with respect to the pivot, located at the base of the boom, approximated here as a uniform beam.) The magnitude of the force applied by the piston is _________ the magnitude of the weight of the boom.
(A) less than.
(B) equal to.
(C) greater than.
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (C)

The weight of the boom acts at its center of gravity, straight downwards. The perpendicular lever arm for the weight force w extends from the pivot to perpendicularly intercept the weight force line of action, such that this will be a horizontal line of length ℓ_w.

The piston force acts straight upwards, and the perpendicular lever arm for the piston force F_piston also extends from the pivot to perpendicularly intercept the piston force line of action, such that this will be also be a horizontal line of slightly shorter length ℓ_piston.

Since the boom is in static equilibrium ("is supported") and does not rotate, then the net torque on it is equal to zero (Στ = 0), such that the clockwise torque of the weight force on the boom must equal the counterclockwise torque of the piston force on the boom:

(cw) τ_w = (ccw) τ_piston.

Substituting in the perpendicular lever arms for these torques:

w·ℓ_w = F_piston·ℓ_piston,

and since ℓ_w > ℓ_piston, then w < F_piston, and thus the force applied by the piston is greater than the magnitude of the weight of the boom.

Sections 70854, 70855, 73320
Exam code: quiz05b0oM
(A) : 7 students
(B) : 24 students
(C) : 24 students
(D) : 0 students

Success level: 42%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.74

Physics quiz archive: rotations, torque, pressure, buoyancy, fluid flow

Physics 205A Quiz 5, fall semester 2016
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, 73320, version 1
Exam code: quiz05b0oM



Sections 70854, 70855, 73320 results

0- 6 :	*** [low = 6]
7-12 :	***********
13-18 :	************************* [mean = 16.6 +/- 6.0]
19-24 :	************
25-30 :	* [high = 30]

Online reading assignment: torque and rotations

Physics 205A, fall semester 2016
Cuesta College, San Luis Obispo, CA

Students have a weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters and previewing a presentation on torque and rotations.


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"Torque is the product of the force and the length of the level arm. The level arm is perpendicular to the line of action (force vector) and the axis of rotation. When the force produces a clockwise direction the torque is negative. Counterclockwise will yield a positive torque."

"How to interpret the line of action and how to draw it."

"Newton has first and second laws applying to rotational motion. These laws are looking at the summation of clockwise or counterclockwise forces on objects."

"If all the counterclockwise and clockwise torque forces balance out then Newton's first law applies. If the forces do not balance out then Newton's second law applies. In this case you can calculate net torque by adding counterclockwise torque with clockwise torque."

"I understand that when under constant motion, the torque will equal zero, and also when stationary. The torque will be nonzero when the forces are unbalanced. I also understand that counterclockwise is positive and clockwise is negative."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"How a lever arm starts and how it intersects the line of action."

"I don't know if the lever arm is measured according to where the force is applied on the object or if its just the length of the entire object."

"I read this section thoroughly and understand it pretty well I'm not confused about anything. This is two to three sentences, can i still get credit?"

What is the SI (Système International) unit for torque?

"A newton-meter. N·m."

"N·m, not to be thought of as joules."

"The SI unit for torque is joules?"

Briefly describe how the line of action should be drawn for a given force.

"A line along the path the force is applied."

"As an extended line drawn colinear with the force."

"It should go into the same direction as the applied force."

"The line of action should be drawn by a straight line?"

"Perpendicular to the force and coming off of the fulcrum?"

"Downward?"

When a lever arm (or moment arm) is drawn, briefly explain where it starts, and how it should intersect the line of action for a force.

"The level arm is perpendicular to the line of action (force vector) and the axis of rotation."

"The lever arm starts at the rotation axis and is perpendicular to the line of the force."

"The lever arm should intersect the line of action at a 90° angle."

"It starts at the center of gravity?"

"It starts and ends at a radius from the point of contact?"

"I didn't understand this."

For the stuck wrench (assuming that it is rigid and not flexing), Newton's rotational __________ law applies, and the clockwise torques and counterclockwise torques acting on the wrench are __________.

first; balanced.	************************************ [36]
second; unbalanced.	** [2]
(Unsure/lost/guessing/help!)	****** [6]

For the crane, Newton's rotational __________ law applies, and the clockwise torques and counterclockwise torques acting on the crane are __________.

first; balanced.	** [2]
second; unbalanced.	*********************************** [35]
(Unsure/lost/guessing/help!)	******* [7]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"I have dealt with torque in calculus and auto classes. I am hoping that helps with executing the physics part of it."

"I am confused with the clockwise and counterclockwise torques and how they can be balanced or unbalanced."

"Why should torque be newton-meters, and not be thought of in units of joules?" (Mathematically torque is a "cross-product" of a force and lever arm, which is different than work, which is a "dot-product" of force and distance traveled. We won't discuss the distinction between cross-product and dot-product operations in this course, but that's why the units of torque and work aren't equivalent.)

"Shouldn't the weight of the objects be taken into account if the fulcrum isn't at the center?" (Yes. Some examples of this in class.)

"Counterclockwise is positive and clockwise is negative--this just seems weird to me." (It's how the unit circle works from trigonometry?)

"How can I get an 'A' in the class?" (You need to solve a math problem: "How many points do I need to get to 600/700 at the end of the semester?")